help-39

[midnight haven]

I don't know what to do

[chrome plank]

how can I get rid of the 1/8 coefficient for x?

[midnight haven]

not really sure

[chrome plank]

well, I can't really add anything. So, perhaps multiplying will work?

[midnight haven]

by what though

[chrome plank]

yeah, or maybe dividing. for example, if I divide 2x = 4 by 2, I can get rid of the coefficient for x. (2x = 4) / 2 is x = 2.

Ok, so what is it particularly about 1/8 that we want to change?

[midnight haven]

the 1

I think

what we do is multiply by 8

[chrome plank]

lets try that

[midnight haven]

both sides?

[chrome plank]

yep. the golden rule is "whatever you do to one side, you must also do to the other side"

[midnight haven]

x+8y=-3

that is an option

[chrome plank]

yep, looks good. does that fit Ax + By = C?

[midnight haven]

yeah it worked

thanks

[chrome plank]

nice

[midnight haven]

.close

[pearl pond]

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[wooden tartan]

could anyone show me a step by step as to how to do this ;-;

[woeful stump]

what's the question

[wooden tartan]

SOrRY it was to find the discontinuitites (if there is any)

[woeful stump]

the questions i would ask are:

• does tan(x) have any discontinuities? are they in the relevant range?
• does your piecewise function have any discontinuities on its boundaries?

[wooden tartan]

but what about the pix/4 next to the tan and the absolute value surrounding the x? what do those do to the problem solving process

[woeful stump]

|x| < 1 just means -1 < x < 1

[wooden tartan]

how does it reacch -1 if the absoute value of x is always positive?

[woeful stump]

ex. -2 is out of the range because |-2| < 1

[wooden tartan]

isnt I-2I 2 which is greater than 1?

[woeful stump]

yeah that's what i meant by "it's out of the range"

|-2| < 1 is not true

[wooden tartan]

OHHHHH

OKAAAY

THANK YOU SO MUCH

SOSOOSOSO MUCH

.close

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[pearl pond]

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[hybrid basin]

Hello

[hybrid basin]

$2\left|x\right|+3\left|y\right|\le6$

[jolly parrot]

Chaewon

[hybrid basin]

How do I graph this function?

[mild socket]

well

this isnt actually a function, is it?

its just an inequality

[hybrid basin]

Oh yeah

Inequality

[mild socket]

so its gonna look like a shaded region

yeah lets figure out how to graph it

ok so if i replaced x with -x would it change anything?

[hybrid basin]

No

[mild socket]

and if i replaced y with -y

[hybrid basin]

No

[mild socket]

right

so lets just focus on when x and y are positive, and we can replicate what we already graphed for the other 3 quadrants

does that make sense

[hybrid basin]

Yes

[mild socket]

alright cool

so lets assume that x and y are positive

[hybrid basin]

Oh wait

Should I just find x-int and y-int

[mild socket]

yes

[hybrid basin]

You are genius

Thank you

[mild socket]

i think you are the genius considering you thought of that lol

just think about where it intercepts the axis, and then connect those points to form a parallelogram

it looks like a diamond

as my 4th grade self would refer to it as

[hybrid basin]

???

4th grade?

[mild socket]

nevermind

no im not in 4th grade lol

but in 4th grade i would have called it a diamond

[hybrid basin]

Oh lol

I think it was called rhombus or something

[mild socket]

yeah it is a rhombus

[hybrid basin]

You have a bright future

[mild socket]

i didnt realize that lol but by the symmetry it is

oh thanks

[hybrid basin]

Teaching a calc 3 victim

[mild socket]

youre in multi?

nice

im in BC

[hybrid basin]

Multi and ODE

[hybrid basin]

Like Canada?

[mild socket]

BC like ap calc bc

[hybrid basin]

Ohhh

[mild socket]

but i alr know all of it

[hybrid basin]

Damn

[mild socket]

so boring

[hybrid basin]

???

You know multi

When you're doing BC?

[mild socket]

oh no

i dont know multi

i know BC

[hybrid basin]

Ohhh

[mild socket]

i know some multi

[hybrid basin]

You would have been cracked

[mild socket]

like surface integrals, line integrals

roughly

i do competition math mostly

[hybrid basin]

Multi should be easy for you if you understand BC well

[mild socket]

yeah

i wish i could take multi this year but BC is the highest a sophomore can take in high school

[hybrid basin]

What do you think about the euclid math contest lol

[hybrid basin]

Mine too

[mild socket]

uh i havent heard of euclid

like the competition

i know the guy

not personally

but like

yknow

i do MAA

amc, aime, e.t.c

and i also do college competitions

[hybrid basin]

Oh god AMC

[hybrid basin]

You should try

[mild socket]

hmm someone sounds traumatized

[hybrid basin]

It's a good comp practice

[mild socket]

alr ill research it

thank you very much

i appreciate it

[hybrid basin]

Np ty also lol

.close

[pearl pond]

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[mild socket]

youre welcome

have a good night/evening/day

[pearl pond]

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[thin cairn]

Hi, I'm struggling with this problem. It wants me to find the value of x. I've tried some things but I can't still found it

[digital pewter]

well, since you don't know what the value of angle B is, cant you put a for it, and then find a?

[thin cairn]

wdym?

[digital pewter]

how would you find a

better question, what do the angles of a triangle add up too

[thin cairn]

180°

[digital pewter]

ok

so knowing that, how would you find a

or which 3 angle measures could you add up to get 180

[thin cairn]

maybe a+30-a plus the unknown value of angle B?

[digital pewter]

yes

and what could you assign for the unknown value

[thin cairn]

could be y?

[digital pewter]

then you would have 30 - a + a + y = 180

and since the a's cancel out, i guess that could work

but then how would you find a

[thin cairn]

that's the point were I'm trapped

'cause a's cancel themselves, so I can't find out the value of a

[digital pewter]

so instead of putting y as the unknown, what could you put instead

[thin cairn]

I'm supposed to replace y with 150, right?

[digital pewter]

no

[thin cairn]

why?

[digital pewter]

you dont know the unknown angle yet

[thin cairn]

the angle B?

[digital pewter]

ya

what variable is being used to represent the unknown angle

[thin cairn]

y

[digital pewter]

oh wait

even that wouldnt work, why does geometry have to be to dumb

sorry about that

[thin cairn]

don't worry

that's the point that confuses me

'cause it wouldn't make sense

[pearl pond]

@thin cairn Has your question been resolved?

[thin cairn]

.close

[pearl pond]

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[half goblet]

Can someone help me with the process of how to factor these 3 equations?

[latent quail]

Let’s look at e

[half goblet]

igh

[latent quail]

Can you factorize this for me?

[half goblet]

just the part u underlined?

[latent quail]

Yes

[half goblet]

ye just gimme a sec

(X+3)(X+3)

[latent quail]

Great

[latent quail]

Can you factorize what I wrote?

[half goblet]

is the second part a z?

-z?

[latent quail]

wdym

[half goblet]

like what is that after the

(X+3

• what

• what

[latent quail]

yes, it’s -z^2

[half goblet]

bru

oh

Ok im not sure how you would factor that

[latent quail]

Do you know (a+b)(a-b)=a^2 -b^2

[half goblet]

conjugates?

i mean its in my notes but

not sure i 100% understand it

ykyk

[latent quail]

I see

You can see x+3 as a whole

[half goblet]

mhm

[latent quail]

It helps when you’re factoring it

[half goblet]

When you say as a whole wdym

[latent quail]

See it as an object

a = x-3 x+3
b = z

[latent quail]

Apply the formula to get the ultimate answer

[half goblet]

it wouldnt happen to be like

(X-3+Z)(X-3-Z)

or something close

[latent quail]

Correct

[half goblet]

LETS GOO

[latent quail]

Ah wait

It’s x+3

Mb

[half goblet]

oh

[latent quail]

Revision

[half goblet]

Ye

I fixed it

Dw

[latent quail]

✅

[half goblet]

Alr run the next one

[latent quail]

First of all, take out 3 and x

[half goblet]

huh

like out the

equation

[latent quail]

No, I mean factorization

For example
2x-4 = 2(x-2)

[half goblet]

Oh

that wouldnt work for 24 though would it

[latent quail]

It would

[half goblet]

but theres no x on the 24

🤔

OH THERES AN X

MB

3x (x-2)(x+4)

Did i do it

[latent quail]

Correct

[half goblet]

IM LOCKED IN

Pretty sure the last one is just difference of squares right

Scratch that

Final question answer is 2(x-2)(x+2)(x^2+4)

atleast thatd what I got

[latent quail]

Correct

[half goblet]

LES GO BABYYYYY

WOOOOOO

[latent quail]

!done

[pearl pond]

If you are done with this channel, please mark your problem as solved by typing .close

[half goblet]

wait I have one more

Just realized

[latent quail]

Alr

[half goblet]

Oh wait

Yea i just realized

Lemme do that

[latent quail]

Alr

[half goblet]

Vouch @latent quail hes the goat

.close

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[hybrid zealot]

Help

[hybrid zealot]

How to solve this , I need to submit before 9 pm

<@&286206848099549185>

Uh

<@&286206848099549185>

[severe hedge]

hi i do not know what the numbers to the left and up of the logs mean

[hybrid zealot]

Fraction

[hybrid zealot]

Can u help?

[severe hedge]

please elaborate on what you mean by fraction

[hybrid zealot]

as you can i see i tried to get the answer by doing eksponen

[severe hedge]

unfortunately i do not understand this

i might have been taught a different way than you

a different helper might be able to help

sorry

[hybrid zealot]

exponent*

[hybrid zealot]

Okay , Np thanks to help

<@&286206848099549185>

[midnight haven]

@hybrid zealot can you explain the notation, like what do those no.s mean on the left of log? Maybe I can help too

I'm not a bot 🙂

[pearl pond]

@hybrid zealot Has your question been resolved?

[midnight haven]

Duudee

[hybrid zealot]

What notarion?

[hybrid zealot]

I think was A root

Root 8 log 4

[pearl pond]

@hybrid zealot Has your question been resolved?

[pearl pond]

@hybrid zealot Has your question been resolved?

[pearl pond]

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[rocky cave]

Can anyone help me with (b) and (c)

Now I partial diff(F) with respect to both x and y to get f
And I try to integrate it back to F but I kinna think that I doing it wrong somewhere

Like this

[pearl pond]

@rocky cave Has your question been resolved?

[rocky cave]

<@&286206848099549185>

[pearl pond]

@rocky cave Has your question been resolved?

[pearl pond]

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[astral grove]

Hello, I'd like help in transforming the marked part into where the arrow points at

[astral grove]

Y is Standard Normal Distributed

I just don't understand why there's the - between the two or how to handle P(value < Y)

[timid citrus]

gyaat

[astral grove]

I do get that we somehow 1 - P(value) and that probably is why the - got dragged inside

[merry carbon]

Everything to the left of sqrt{n}/2, but then forget the stuff to the left of -sqrt{n}/2

[astral grove]

Honestly I'm not even sure at all how to do the conversion ^^° So I'd like either somewhere to look or an explanation for that

[timid citrus]

gyas\

gyaaaaaaaaaaaaaaaaat

[astral grove]

Uhh, sorry I don't understand you?

[merry carbon]

Ignore them

[astral grove]

But yeah idk what this process is called so I can't even look it up or ask for it that's why I'm asking here

[merry carbon]

Do you know what this represents?

[astral grove]

It's a function for a standard normal distribution

You plug in a value and look up in a table the output value

There's some more complicated function hiding behind it I think

[merry carbon]

Or, alternatively, you're happy that 0.1/0.2 is 1/2, right?

[astral grove]

Oh yeah

I'm not asking about the /2 part

I'm asking about what happens to the minuses

[merry carbon]

Cool then - are you happy, that this is true:
[
P(-\frac12 \sqrt{n} < Y < \frac12 \sqrt{n}) = P(Y < \frac12 \sqrt{n}) - P(Y < -\frac12 \sqrt{n})
]
?

[jolly parrot]

@merry carbon

[astral grove]

Uhhhhh I definitely have questions for the minus in the middle

I'm currently thinking about what happened with the P at the start, it looks like the - got consumed to transfer it to the other side

So the 1 is the one I don't understand at all where it's coming from

The 2 I can guess why

[merry carbon]

If you're between two numbers, say a and b, then you're less than b and greater than a, yes?

[astral grove]

Yes

[merry carbon]

So you're less than b, and not less than a, right?

[astral grove]

Oh b - a

Oooooooh

[merry carbon]

Basically that, the first probability is that you're to the left of sqrt{n}/2, which gets us a bit more than we wanted

[astral grove]

Okay, thanks already for that now I get it
How do you do the conversion

[merry carbon]

However, we actually want to be larger than -sqrt{n}/2, so we do not want to be to the left of -sqrt{n}/2, so we remove the probability that we are

And which conversion, to the final form they said?

[astral grove]

To the psi but ig I have to look that up on my end

[merry carbon]

tl;dr if $Z$ is standard normally distributed, then $\Phi(x)$ represents the probability $P(Z < x)$

[jolly parrot]

@merry carbon

[astral grove]

I'm unsure about this because I also saw 1 - P (X_i smaller equal value) = 1 - psi(value)

I found the definition for it

It's the distribution function which makes a lot of sense why we are interested in that

[merry carbon]

Yea, as above it represents the probability that a standard normally distributed variable takes a value at most x

[astral grove]

So is it explicitely <, or =<, or does it not matter

[merry carbon]

It doesn't matter for continuous random variables - probability of being equal to any specific value is always 0

[astral grove]

That makes sense

Thank you a lot

I was really struggling with this ;-;

Can I close it?

[merry carbon]

Yep sure, if you're happy

[astral grove]

Thanks again!

.close

[pearl pond]

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[pearl pond]

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[shy ocean]

Could someone please point out the error here?

[pearl pond]

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[boreal slate]

hey, so i understood the proof of this one:

[boreal slate]

but i didnt manage to understand how to proof for n=4 like here:

[boreal slate]

relying on this proof

[lime river]

They have used the above Identity by taking x = x1+x2/2 and y = y1+y2/2

[boreal slate]

ok the v means understood and ? means not understood

help?

(all x>0)

[lime river]

Use am gm inequality with x = x1+x2/2 and y = x3 +x4/2
So we have x +y /2 >= √xy
x 1 + x2 + x3 + x4 /2 >= √(x1+x2)(x3+x4)
Now square both sides two times to obtain the result

[pearl pond]

@boreal slate Has your question been resolved?

[boreal slate]

how x = x1+x2/2 and y = x3 +x4/2

but u say x 1 + x2 + x3 + x4 /2 >= √(x1+x2)(x3+x4)

sry if im making u hard time..

thats what i did

for x = x1+x2/2 and y = x3 +x4/2

[lime river]

Now square that two times

[lime river]

By observation

[boreal slate]

ok trying

[indigo egret]

hey guys i know this is a math group but can anyone please help me with this ?

[boreal slate]

post it on MATH HELP (AVAILABLE)

[indigo egret]

alright thanks

[boreal slate]

i did square once but i doing it one more time makes no sense to me.. help??

my question is how u obtain that?

[pearl pond]

Someone else is already using this help channel. If you need help with a question, please open your own help channel/thread (see #❓how-to-get-help for instructions).

[lime river]

Ok here

[pearl pond]

@boreal slate Has your question been resolved?

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[boreal slate]

Thank you!!

[pearl pond]

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[stoic imp]

I mean

[west sapphire]

which one are you asking about?

[stoic imp]

a and b

like how can be the same

but at the same time differen

t

[west sapphire]

what do you mean by that

[stoic imp]

sorry I just looked its x/1-x^2

vs 1/1-x^2

I mean in general how do I solve this exercises?

I dont really know what I am doing am I allowed to say that

[west sapphire]

in your original formula i'm going to use a different variable to avoid confusion: $$\sum_{n=0}^\infty u^n = \frac{1}{1-u}$$

[jolly parrot]

Bungo

[west sapphire]

part (a) is just setting u = x^2

[stoic imp]

okay

[lime river]

It's binomial expansion for negetive powers

[stoic imp]

binomial what?

how so? is this related to combinatorrics

[lime river]

No it's related to combinatorics

spell it correctly

[stoic imp]

cab you guys help me with abcd instead

I mean bcd

because a was plugging in x mostly

[west sapphire]

well look at the right hand sides of a and b, how are they related?

[stoic imp]

it's the same stuff with multiply of x

[west sapphire]

yea, and how about the left hand sides

[stoic imp]

same stuff multiplied by an x

[west sapphire]

yea

[lime river]

Well quite many ways to do it
The simplest one for b is forming a gp

[west sapphire]

so take (a) and multiply it all by x

done deal

[stoic imp]

GP?

[lime river]

leave it do as the other guy says multiply by x

[stoic imp]

what about c and d?

[west sapphire]

try the same idea with (c), look at the right hand sides, how are they related?

(compared to the original, not to (a) or (b))

[stoic imp]

first of all I dont know what does the |x| < 1 means overall but, for the c:

it's the same as x^n = 1/1-x but with a (-1)●x as x

[west sapphire]

yea so if we use "u" to avoid confusion

as in

then (c) is taking u = -x

[lime river]

It means that the absolute value of x is less than 1

[west sapphire]

yea the original series only converges if |x| < 1

[lime river]

and it's the most imp detail in the sum as other wise the sum would be diverging

[stoic imp]

my follow-up question would be why LHS of c) is only (-1) for odd

like negative

[west sapphire]

well what's (-x)^2

[lime river]

[stoic imp]

(-1)^2 ● x^2

[west sapphire]

and (-1)^2 = ?

[stoic imp]

1 yeah sorry I am bad

[lime river]

Def not 1

[west sapphire]

and same for all even powers

[summer gorge]

so true

[stoic imp]

I mean tbh this exercise looked scary but if you read it nicely and after you guys explained I think I understand more

[west sapphire]

so what are your thoughts about (d)

[stoic imp]

u = 2x

[west sapphire]

yea

[summer gorge]

correct

[west sapphire]

and what does |u| < 1 translate to

in terms of x

[stoic imp]

2|x| < 1 <==> |x| < 2^-1

[west sapphire]

yep, so that's where the last bit comes from

check my terrible cropping haha

[summer gorge]

its not bad lol

[stoic imp]

which identity is this 1/1-x

[west sapphire]

they didn't prove that for you in your class or book or whatever?

[summer gorge]

GP for infinite series

wait i forgot the exact name

[stoic imp]

geometric progression?

[west sapphire]

$$(1-x)(1 + x + x^2 + \ldots + x^n) = (1 + x + x^2 + \ldots + x^n) - (x + x^2 + \ldots + x^{n+1}) = 1 - x^{n+1}$$

[jolly parrot]

Bungo

[west sapphire]

so dividing both sides by 1-x, you get

$$1 + x + x^2 + \ldots + x^n = \frac{1 - x^{n+1}}{1 - x}$$

[jolly parrot]

Bungo

[lime river]

He isn't familiar with gp or binomial expansion or others and the exercise mentions he has to use identities given

[west sapphire]

now take limits as n->infinity, using the constraint |x| < 1

[stoic imp]

I mean geometric progression was probably introduced, binomial expansion most surely not

[west sapphire]

you don't need binomial expansion here

[stoic imp]

is just an extra exercise also, is optional

[lime river]

Taylors theorem ?

[stoic imp]

yes

[lime river]

Nice

[west sapphire]

just the above calculation, and then some simple substitutions

[stoic imp]

introduced to it

okay ty

.solved

[pearl pond]

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[toxic trout]

What’s wrong here?

[verbal whale]

You have 1 + **(ln(x))**², not **(1+ln(x))**²

So you should sub u = ln(x)

[toxic trout]

Ok thx

[verbal whale]

At first glance I admit I also read it the wrong way lol

[old marsh]

i like the sub u =1 + (ln(x))²

[verbal whale]

I would do it in two steps, first u = ln(x), and then t = 1+u²

However, that's it yeah

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[spice condor]

For a function f: R→R, let C denote the shape expressed by the equation
(y-f(x)) (y-x³-9x² - 24x) = 0
in the Cartesian plane. For each real t, let g(t) be the number of distinct inter- sections between C and the line y = t. Let F denote the set of all linear functions f such that there are at least two distinct values of k satisfying
g(k)=|lim t->k- g(t) - lim t -> k+ g(t)|
Find the maximum and minimum of f (6) for f € F.

[pearl pond]

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[torpid pelican]

need help with this i have worked on it but thinking of getting a more efficient answer. it is a puzzle from discrete math class. related to predicates equivalence and inferences.

[regal herald]

most people dont like to open files, youre better off sending an image

[full relic]

What have you tried

[midnight haven]

The Numbers corespond to the number of bombs beside them up down left and right the corners also count

Can I say the answers or do you want to figure it out?

[full relic]

Do not give solutions

[midnight haven]

Sir, yes sir

Opps

[pearl pond]

As a helper, please do not give out answers that could be copied as a homework solution. Have the student work through the problem themselves and guide them along the way.

[torpid pelican]

smthin like this

haven't reached to the solution but you get the idea right?

[midnight haven]

A1 doesnt have a square nor number, should prob put somn to avoid confusion

[torpid pelican]

so shouldn't include A1?

[feral leaf]

Are you using the x's to indicate there's a mine?

[torpid pelican]

ye

[feral leaf]

You have 6 x's, the puzzle statement said there are 6 mines hidden

[midnight haven]

Black it out maybe? Just somn to indicate its not a square thingy

[torpid pelican]

ye 6 x = 6 mines

found them

[feral leaf]

So then what does this statement indicate?

[torpid pelican]

ok

[feral leaf]

Because that doesn't match with what you drew in the corner

[torpid pelican]

according to the problem i have to answer in a logical statement

[torpid pelican]

it doesn't?

[feral leaf]

The x's are mines

Why did you write A1 has a mine when you did not put an x there

[torpid pelican]

i m trying to convert it into logical statement as i mentioned "exactly one must be true (having a mine)"

it doesn't have to be A1

[feral leaf]

The confusion is you drew this, but then listed those 5 spots, they don't match up

[torpid pelican]

but A1 should be excluded so its wrong ofc

[midnight haven]

Uhm A1 isnt a square its just a blank slate so it isnt part of the probabilities, ye

[torpid pelican]

i just wrote all the adjacent ones and add OR operator so one must have a mine

i m so confused and deadline is in 3 hrs

if i remove A1 and start like this having OR operator

[midnight haven]

In the photo you sent A1 is just a blank slate so it cant be a mine

[torpid pelican]

ye

now ik that (bcs of u)

[midnight haven]

Soooo A1 cannot be a possibility

[torpid pelican]

where exactly one must be true

[midnight haven]

Whats that M thingy mean?

[torpid pelican]

i let M(x, y) = there's a mine at (x, y)

[midnight haven]

Why is there a mine at A3?

[torpid pelican]

there isnt

just to create logical statement

and then create same for other tiles

combining the constraints

thats what i thought

what do you think about this?

[midnight haven]

Sooo youre saying theres a mine aroundd it?

[torpid pelican]

1 must be true means exactly one mine around it, ye

[midnight haven]

Dis is some hard stuff, I like simplifying things and this... Is not simple😥

But I kinda get it

[torpid pelican]

now what should i do :((

[midnight haven]

Does the proof have to be equations?

[torpid pelican]

logical statement

involving predicates, inferences and equivalence rules

[midnight haven]

Well what you did with this one looks good

[torpid pelican]

it does?

but how can i know i m going in the correct direction

[midnight haven]

Its basicly just circling around the number until you hit a tile but yeah

[torpid pelican]

i see

what if i ping helpers and sm1 comes and see it?

[midnight haven]

You should probably use the other numbers near the tile too to add more reason why this tile has to be a mine

[torpid pelican]

ye repeating this for all numbered tiles

[midnight haven]

Welp good luck too you bro

[torpid pelican]

thenks

<@&286206848099549185> anyone wants to look at the problem and help b4 i close this :((

[midnight haven]

Also did you add somn to A1 to indocate thay its a blank slate? Dont want you getting confused

[torpid pelican]

i coloured that tile showing that its blank

[pearl pond]

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[pulsar stump]

why can we use this formula det A? we didnt prove it the same as the premutation formula

can someone write the proof or send me a link to one

[plush bramble]

Google cofactor expansion

[pulsar stump]

no proof

[cinder flower]

google cofactor expansion theorem proof

https://math.emory.edu/~lchen41/teaching/2020_Fall/Section_3-6.pdf this for example comes up for me

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[obtuse flame]

The question is to find all values for x in the equation.

[obtuse flame]

I've managed to find one of the values. However, the book also suggests that x = n is a solution. I do not know where that came from.

[long gorge]

can you show your work?

[obtuse flame]

The work is above. The first line is the equation itself, whilst the rest is my work

[long gorge]

oh sorry 🤦

so you have [\qty(\frac{x}{n})^{\lg x} = \qty(\frac{x}{n})^{2}]

[jolly parrot]

Invariance

[long gorge]

one way that can be solved is lg x = 2

but could it possibly be true in any other way when lg x isn't 2?

[obtuse flame]

it could be true if lg x = lg n for any value of n, according to the book

[rough stream]

On line 2, take the lg of both sides, drop the power. Then, solve for lg(x).

[obtuse flame]

I do understand this now. Thanks

.close

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[long gorge]

the other way it could be true is if x/n is 1

[long gorge]

because 1^x = 1^y for any x and y

and if c^x = c^y but x != y, then c = 1 or -1 (if c is a real number)

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[autumn cliff]

Hiii! I’d just like to check if my sketching of this graph is correct (the limits and points I had to sketch the graph to follow are on the side).

PS: I gotta run to my next class eeee but I'll check back here after that. thank you so much <33333

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[flat stirrup]

I would like to ask some stuff about projective geometry. Suppose we have 6 Points A,B,C,D,E,X. And A,B,C,D are on a conic section. And we know that (A,C;B,D) = -1 from the perspective of E. Now I have to construct with harmonic conjugate to show that X is not on the conic section...

[flat stirrup]

So I know that the cross ratio of (A,C;B,D) from the perspective of E and (A,C;B,D) from the perspective of X should be both -1 if X is on the conic section

[glass salmon]

whats X?

[flat stirrup]

X is also a point

[glass salmon]

wait what are you looking to find?

[flat stirrup]

i am suppose to use harmonic conjugate to construct

so that i can show

[glass salmon]

harmonic conjugate of which 3 points

[flat stirrup]

that X is not on the conic section

[glass salmon]

the locus of points such that P(A,C;B,D)=-1 is the conic through ABCDE

[glass salmon]

what if i choose X to be on the conic section?

[glass salmon]

then this claim is wrong

[flat stirrup]

the position of the points are given

[glass salmon]

ok then to do it geometrically

choose a line parallel to XA

and project (A,C;B,D) onto it from X

let XC, XB, XD intersect this line at P,Q,R

then manually check if the midpoint of PR is Q

if so, then X lies on the conic

if not, X does not

[flat stirrup]

ahh ok let me check

[flat stirrup]

im not sure how to do that...

[glass salmon]

It Means intersect XA, XB, XC, XD with the line

XA will be parallel to the line so XA intersects it at infinity

[flat stirrup]

so the midpoint of PR is not Q

but how does this work?

[glass salmon]

try it

hint: use length definition of cross ratio

(with the point at infinity which is XA intersection with the line)

[flat stirrup]

so we dont need E here at all per se?

[glass salmon]

no we do

because E tells us for any point P on the conic through ABCDE, P(A,C;B,D)=-1

[flat stirrup]

ok so now i have this

and if X is on the conic

then (P, infinity; R, Q ) = -1 ?

[glass salmon]

Yes

and this implies P is midpoint of QR

[flat stirrup]

ohhhh

now i get it

Many thanks!!

[glass salmon]

np

extension may be to let E(A,C;B,D)=x for some real x and figure out how to determine if X lies on conci through ABCDE, but its hte same solution

so harmonic wasnt completely necessary

[flat stirrup]

oh. if x is not -1 then P mussnt be the midpoint then right?

[glass salmon]

Yeah, it will be another ratio

[flat stirrup]

ok!

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[plush bramble]

use ratio test

[plush bramble]

did you take the ratio ?

[stoic imp]

on it

a_n on the numerator or denomi

?

[pearl pond]

@stoic imp Has your question been resolved?

[stoic imp]

.

<@&286206848099549185>

[plush bramble]

What are you stuck on

[stoic imp]

how do I find the radius of convergence

[stoic imp]

.

[pearl pond]

@stoic imp Has your question been resolved?

[plush bramble]

Last two lines are wrong

The x doesn't become 0 magically

[stoic imp]

okay how would you do it

[plush bramble]

https://m.youtube.com/watch?v=iI_2Piwn_og

[pearl pond]

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[sudden pawn]

I need help 😭😭

[sudden pawn]

That's not the right orientation

Err

[sudden pawn]

Nvm

Found it

.close

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[abstract halo]

for x1 I have dx/dt = -5x/3 -> x1(t) = Ce^(-5t/3)

for x2 I have dx_2/dt + 5/3x_2(t) = 5/3 x_1(t)

-> x_2(t) = x_1(t) + Ce^(-5t/3)

and for x_2, C = -x_1(t)

how can I find for c?

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[pulsar stump]

why is he allowed to do the bottom steps of extracting scalars (lambda_i-lambda_1) from Minor_1,1?

[pulsar stump]

it's wrong

he can only take inverse of (lambda_i-lambda_1)

[pearl pond]

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[red falcon]

let's say S_n(x) is a sequence of partial sums of a series of functions

we know that if $S_n(x) \to S(x)$ uniformly as $n \to \infty$ then the $\sum f_n(x)$ converges uniformly to S(x)

[jolly parrot]

atif

[red falcon]

but to me it seems that here(in the image above) is being said that, since S_n(x) doens't converge to S(x) (and it doesn't bcs 1=/=0) the series doesn't not converge uniformly to S(x)

am I right in thinking that we can't claim this? since we have only $\implies$ and not $\iff$

[jolly parrot]

atif

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[winter moat]

is there a possible was to use the right triangle ratios

way*

im mostly confused and stuck at drawing the triangle

idk how to draw the triangle correctly

wait i think i got it

ok

.close

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[atomic hound]

ok so this is what im dealing with

and the problem is that i dont understand whats happening

could someone please explain

i cant even find this on the internet so this is my last hope

[plucky python]

ok so we're basically proving the 'obvious' statement that if you give me 2 numbers

i can divide 1 by the other to get a unique remainder

so restated in terms of symbols, that means that for any $n,m$, there exists unique $q, r$ such that $n=qm+r$

[jolly parrot]

LY

[plucky python]

(0 <= r < m)

[plucky python]

also you should be able to find this on the internet

it's just called like division algorithm/euclidean division

[plucky python]

anyway back to this

the statement has 2 parts: existence and uniqueness

so we need to prove that we can actually divide 1 number by another (which should be obvious)

and that we can get a unique remainder (which should be obvious again if u've ever divided 1 number by another)

lemme read what the proof u've been given does

[plucky python]

yeah ok so which part of the proof don't u understand?

[atomic hound]

where does the P come from in the third line like why p

[plucky python]

that's just a 'dummy variable'

it's just how we construct our set

[atomic hound]

okok

[plucky python]

i think it might help to have a numerical example

so say i give you 17 and 5

we want to show that we can divide 17 by 5 to get a remainder

how would you do it?

[atomic hound]

like 15/5=3 and 2/5 is the remainder

everything makes much more sense after ur explanation tho

[pearl pond]

@atomic hound Has your question been resolved?

[plucky python]

nw!

[plucky python]

yeah exactly - you pick 3 because

3*5 < 17

but 4*5 > 17

that's all the first part is doing

and then for the 2nd part (uniqueness), basically it's just

if you pretend that 17 can have 2 different remainders

say 2 and 4

then 4-2 must be a multiple of 5

which is absurd

[atomic hound]

why must 4-2 be a multiple of 5 tho

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.reopen

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[flat stirrup]

a question in projective geometry

[flat stirrup]

so suppose we have a projective transformation in CP1 which is complex

and we know the projective transformation which is T

how can we interpret it geometically with a complex point z= x + iy?

T = (2i, 4 ; 0 ,2) as a 2x2 matrix for example

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.reopen

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✅

@flat stirrup Has your question been resolved?

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@flat stirrup Has your question been resolved?

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@flat stirrup Has your question been resolved?

[pearl pond]

@flat stirrup Has your question been resolved?

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[glossy roost]

am i trippin

6.1 to 7.9s looks right

<@&286206848099549185>

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@glossy roost Has your question been resolved?

[tulip ore]

yea it is

[pearl pond]

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[stuck elbow]

i'm working with bessel functions, and the problem statement begins with,

In a problem with circular symmetry, let $f(\rho)$ be a function describing the initial profile of a function obeying Bessel's equation with boundary conditions $J_m(ka)=f(a)=0$ for some $m$. We write
$$f(\rho)=\sum^\infty_{n=1}c_{mn}J_m(k_{m,n}\rho)$$
I won't get into the specifics of the problem, but I was wondering whether I was allowed to use the method of frobenius to expand $f(\rho)$. Thanks!

[jolly parrot]

arizona

[full relic]

Remind me again which one is Frobenius

Though uhh can’t say I’m too familiar with this particular equation, but the most obvious concern would be like, regularity or convergence yeah?

[pearl pond]

@stuck elbow Has your question been resolved?

[pearl pond]

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[robust oar]

what have you tried

[bold seal]

4 x 2

2x 2

No wait

I meant

Is that I have no clue

[robust oar]

you need to apply the log rules

[bold seal]

What are those?

[robust oar]

,tex .log rules

[jolly parrot]

[merry carbon]

(irrelevant but didn't know the bot does emotes even from other servers )

[robust oar]

it is a secret

[bold seal]

So we just put the 4 behind log?

[acoustic path]

[bold seal]

So it would be 4log2 2^x?

[robust oar]

[robust oar]

not quite

[bold seal]

Multiply the 2x?

To be 16x?

[robust oar]

no

[bold seal]

Then what should I do?

[robust oar]

you need to apply the product rule first and then after this

you need to solve the log and rewrite it nicely

or if you want you can use this rule $\log_b (m^n) = n \log_b m$

[jolly parrot]

[robust oar]

same thing as power rule

[bold seal]

So would it not be 4log2 2x?

[robust oar]

do you mean $4log_2 2^x$? then no

[jolly parrot]

[merry carbon]

(think they meant $4\log_2(2x)$, in which case, $(2x)^4$ vs $2x^4$)

[robust oar]

cursed log

[jolly parrot]

@merry carbon

[robust oar]

ic

[merry carbon]

(also so the emote thing does work, at least from any server the bot's in )

[robust oar]

ye

but maybe it doesn't if my nitro is gone

lmao

[bold seal]

I’m not sure how to solve. This

4log2 x?

[robust oar]

no

[robust oar]

ill restart again

what will happen if you apply the product rule

write it down

[bold seal]

log (2) + log (4) ?

[robust oar]

close

but you forgot the base

[bold seal]

log 2(2) + log 2(4)?

[robust oar]

do you mean this $log_2(2) + log_2(4)$

then no

[jolly parrot]

[robust oar]

you forgot something

[bold seal]

multiplying?

[robust oar]

no, try again for the second one

log_2 2 is correct

but the other isn't

[bold seal]

log (4)?

[robust oar]

not quite

try again

[bold seal]

im not quite sure where to go from here

[robust oar]

you have two different variables lets just assume the log_2 2 is x and log_2 x^4 is y

you got the x is correct but for y almost

[bold seal]

log 2(16)?

[robust oar]

no

[merry carbon]

If you were given $\log_2(2q)$, would you know how to deal with that?

[jolly parrot]

@merry carbon

[bold seal]

solve q?

[merry carbon]

Expand that expression

[bold seal]

log_2 q

[merry carbon]

Remember that you have the log rules as per here

[bold seal]

2 log_2q?

or 2log2

[merry carbon]

You're happy that 2q is 2 * q, right?

[bold seal]

im not sure how to do this.

[merry carbon]

Are you happy to accept, that the log rules here, tell you that $\log_a(pq) = \log_a(p) + \log_a(q)$?

[jolly parrot]

@merry carbon

[pearl pond]

@bold seal Has your question been resolved?

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[fleet magnet]

I have to describe a graph model.

[versed mica]

not funny man

[fleet magnet]

hello

I don't know how I would model this question so that each family is connected to each other to form a graph.

would I assume each children is their own family?

[pearl pond]

@fleet magnet Has your question been resolved?

[pearl pond]

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[dire snow]

Can’t find an integrating factor that works

[acoustic path]

x^-2 works

[dire snow]

ty

[pearl pond]

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[fleet magnet]

I don't know how to approach this problem.

[pearl pond]

@fleet magnet Has your question been resolved?

[ocean hornet]

i think what you can do is take two vertices in two different connected components, connect them so that you don't get another face, and repeat until you have a connected graph

[fleet magnet]

This is for a)?

[ocean hornet]

yeah

this is a bit hand wavy because you are assuming that there are vertices in connected components that don't give you another face when connected

[fleet magnet]

let me see

[ocean hornet]

but if you have C connected components, you have C-1 extra edges you need to add to make the entire graph connected, hence v-(e+c-1)+f=2

[fleet magnet]

what is a face?

is it the same thing as a region

[ocean hornet]

err yeah

i've never heard it called a region

the space bounded by the edges and vertices

[fleet magnet]

I'm confused. Why do you have c-1 extra edges to make the entire graph connected?

[ocean hornet]

let's say you have 2 connected components, you only need 1 edge to connect them

similarly, if you have c components, adding one edge will reduce the number of components by 1, so we need c-1 edges to reduce to 1 component

[fleet magnet]

so you just added c-1 + the number of edges in all the componenets combined?

[ocean hornet]

well that would be the total number of edges in the resulting graph

and since it is a connected graph, we can plug the numbers into euler's formula

so v-e+f=2 -> v-(e+c-1)+f=2

[fleet magnet]

wouldn't you be able to add 2 extra edges for each component because faces need to be bounded by at least 3 edges?

[ocean hornet]

err, we conceivably could, but why would you want to? we just need to make the graph connected so that we can apply euler's formula

[fleet magnet]

oh ok

then for b) what do I have to do?

[ocean hornet]

actually i just thought of a simpler way that gives the same answer, let the vertices, edges, and faces (exlcuding the outside face) of the connected components be v_i, e_i, and f_i, then we know that v_i - e_i + (f_i+1) = 2, now we can sum over all c components to get v - e + f + c + 1 = 2c

gives the same answer

[ocean hornet]

is it just asking for a proof?

im not sure what it's asking

[fleet magnet]

I think it's asking me to show that e <= 3v-6 also holds for unconnected

[ocean hornet]

oh i see

[fleet magnet]

do I still use 3f >= 2e and plug it in?

[ocean hornet]

err yes you should be able to

because you have that e <= e+c-1 <= 3v-6

[fleet magnet]

how is e + c -1 <= 3v-6?

[ocean hornet]

well in the connected graph with the extra c-1 edges, we can apply euler's formula which implies the corallary, so since the number of edges is e+c-1, the corollary says e+c-1<=3v-6

[fleet magnet]

I got:
e <= 3v -3c -3
Is it fine to say that because c >= 1 then e <= 3v -3c -3 <= 3v-6?

[ocean hornet]

oh whoops i meant extra c-1 edges

not vertices

so E <= 3v-6, and E = e+c-1, so e <= e + c-1 = E <= 3v-6

[fleet magnet]

I have a stupid question: if you have an expression like -x/-(x-2) < y and you cancel out the -1, do you have to flip the inequality as well?

[ocean hornet]

no, because -x/-(x-2) = x/(x-2)

if you multiply on both sides of the inequality by -1, then you need to flip

[fleet magnet]

then what am I doing wrong here:

[ocean hornet]

what is k

also what the heck is going on

[fleet magnet]

it's for this problem, I did kr <= 2e

[ocean hornet]

oh well it seems like you divided by 2-k

which is negative probably

that reversed the inequality

[fleet magnet]

oh because k >= 3

ok I have to watch out for that

[lucid marsh]

hey

[lucid marsh]

anyone got this yet?

[pearl pond]

@fleet magnet Has your question been resolved?

[pearl pond]

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[lavish ember]

When to use ± sign when solving a geometric mean?

[acoustic path]

never, geometric mean is defined to be positive

[lavish ember]

hmm wait, maybe i asked the wrong question.... insert a geometric mean between k and 1/k?

thats the question im answering, but after getting the GM of 1, i realized it can either be 1 or -1 right?

[acoustic path]

.

[lavish ember]

oh

alright

thanks for the infoo

.close

[pearl pond]

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[sweet cedar]

unsure how to do 1a

[sweet cedar]

just need help with it if someone could explain how to do it i could probably do 1b

[pearl pond]

@sweet cedar Has your question been resolved?

[red sky]

alr